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(C) The average power dissipated in an $a.c.$ circuit is given by the formula: $P_{	ext{avg}} = V_{	ext{rms}} I_{	ext{rms}} \cos(\phi)$.Given $V =

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$2.5$

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(C) The average power dissipated in an $a.c.$ circuit is given by the formula: $P_{	ext{avg}} = V_{	ext{rms}} I_{	ext{rms}} \cos(\phi)$.Given $V = 100 \sin(100t) \ V$,the peak voltage $V_0 = 100 \ V$.Given $I = 100 \sin(100t + \frac{\pi}{3}) \ mA$,the peak current $I_0 = 100 \ mA = 100 	imes 10^{-3} \ A = 0.1 \ A$.The phase difference $\phi = \frac{\pi}{3}$.$V_{	ext{rms}} = \frac{V_0}{\sqrt{2}} = \frac{100}{\sqrt{2}} \ V$.$I_{	ext{rms}} = \frac{I_0}{\sqrt{2}} = \frac{0.1}{\sqrt{2}} \ A$.$P_{	ext{avg}} = \left(\frac{100}{\sqrt{2}}ight) 	imes \left(\frac{0.1}{\sqrt{2}}ight) 	imes \cos\left(\frac{\pi}{3}ight)$.$P_{	ext{avg}} = \frac{100 	imes 0.1}{2} 	imes \frac{1}{2} = \frac{10}{4} = 2.5 \ W$.

In an $a.c.$ circuit,voltage and current are given by: $V=100 \sin (100 t) \ V$ and $I=100 \sin (100 t + \frac{\pi}{3}) \ mA$ respectively. The average power dissipated in one cycle is: (in $W$)

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